Why is the Volume Distribution of Charge Described Differently in the Book?

Ripperbat
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Homework Statement



A rod of dielectric material is spun about it's axis with angular velocity \omega. A uniform magnetic field B exists in a direction along the axis of the bar. Determine a charge distribution which produces the same electric field as does the rotating rod. The electric susceptibility of the material is xE

Homework Equations





The Attempt at a Solution



The force on a charge q at a distance r is F=q*\omega*r*B and the electric field is E=r*B*\omega.
The polarization of the rod at distance r is P=xE*\epsilon0*\omega*B*r.
So the volume distribution of charge should be -div P = -(\partialPr) / (\partial*r). But according to the book it should be
-div P = -(\partialPr) / (\partial*r) - (Pr) / r .


Could somebody please explain why it should be that way?? thx
 
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Ripperbat said:
So the volume distribution of charge should be -div P = -(\partialPr) / (\partial*r). But according to the book it should be
-div P = -(\partialPr) / (\partial*r) - (Pr) / r .


Could somebody please explain why it should be that way?? thx
That is the divergence in cylindrical coordinates.

http://hyperphysics.phy-astr.gsu.edu/Hbase/diverg.html#c3
 
Thank you thank you thank you! :!)
 

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